Surface $S$ is parametrized by $$X(u,v) = (r (v) \cos{(u)}, r(v) \sin{(u)}, z{(v}))$$ with everywhere-constant Gaussian curvature $K$. Rotation $u \in (0, 2 \pi)$. $v$ is the arc length of the generating curve $(r(v), z(v))$ of $S$; primed with respect to arc $v$ thus for unit speed we have
$$ (r')^{2} + (z')^{2} = 1\tag 1 $$ Prove that $r$ satisfies $$ r'' +K r = 0\tag 2$$
and
$$z= \int \sqrt{1-(r')^{2}} dv\tag 3 $$
where the domain of $v$ is such that this integral makes sense.
Let us fix a plane with origin $O$ and directions $x$, $y$; denote this plane $\langle x,y \rangle$ by name $xOy$.
Fix $K=1$ positive and let $S$ intersect $xOy$ perpendicularly. Prove that, for some $C = r(0) \in \mathbb{R}$,
$$r(v) = C \cos{(v)}, z(v) = \int_{0}^{v} (\sqrt{1-C^{2} \sin{(v)}^{2}}) dv\tag 4 $$
Next show that all surfaces $S$ of revolution of fixed everywhere negative constant Gaussian curvature $K=-1$ and parametrized by a function $X$ of $u$ and $v$ (in that order), are described by the following (up to Euclidean motion):
$$ (a:) \, r(v) = C \cosh{(v)}, z(v) = \int_{0}^{v} (\sqrt{1-(C \sinh{(v)})^{2}}) dv \tag 5$$
$$ (b:) \, r(v) = C \sinh{(v)}, z(v) = \int_{0}^{v} (\sqrt{1-(C \cosh{(v)})^{2}}) dv \tag 6$$
$$ (c:) r(v) = \mathbb{e}^{v}, z(v) = \int_{0}^{v} (\sqrt{1-\mathbb{e}^{2v}}) dv \tag 7$$
I am thinking differential equations, but I am not sure how to set them up.
Calculating the coefficients of the first and second fundamental forms we have:
$$E = \varphi^2$$ $$F = 0$$ $$G = (\varphi')^2 + (\psi')^2 = 1$$
and the second form:
$$L = -\varphi\psi'$$ $$M = 0$$ $$N = \varphi' \psi'' - \psi'\varphi''$$
Derivating $ (\varphi′)^2 + (\psi′)^2 = 1$. We have $\varphi'\varphi'' + \psi'\psi'' = 0$
Thus, $N$ can be written as $ N = \varphi''/\psi'$.
Finally, being $K = \frac{LN - M²}{EG - F^2}$
$$K = \frac{LN}{EG} = \frac{-\varphi\varphi''}{\varphi^2} = - \frac{\varphi''}{\varphi}$$
Then we get, $$ K\varphi + \varphi'' = 0$$
$$Q.E.D.$$